This cubic is a nodal K60+ (or a stelloid) with a node at H where the tangents are parallel to the asymptotes of the Jerabek hyperbola. The tangents at A, B, C are the symmedians.

The three asymptotes concur at X(381), the midpoint of X(2)X(4), and are parallel to those of the McCay cubic.

It is (partially) described in Musselman's paper (see bibliography) in the following manner. Denote by A2, B2, C2 the reflections of P in the sidelines BC, CA, AB and by A3, B3, C3 the reflections of H in the lines AP, BP, CP. The three circles AB2C2, BC2A2, CA2B2 have a common point N on the circumcircle. The three circles PAA3, PBB3, PCC3 have P in common and another point L also on the circumcircle. These points N, L coincide if and only if P lies on the Musselman (second) cubic K027 and they are antipodes if and only if P lies on the Musselman (third) cubic K028 (Musselman, Some loci connected with a triangle. Monthly, p.354-361, June-July 1940).

K028 is also the isogonal transform of the Lemoine cubic K009, the isotomic transform of K257 and the anticomplement of K258. Note that K009 and K028 define a pencil of cubics that also contains K005 and K187. The X(1989)-isoconjugate of K028 is the nodal stelloid K724.

Another description is the following : let P be a point and Pa, Pb, Pc its projections on the perpendicular bisectors of ABC. Triangles ABC and PaPbPc are perspective if and only if P lies on the Stammler hyperbola. The locus of the perspector is the Musselman (third) cubic.

K028 is the Phi-transform of the circumcircle and the Psi-transform of the line at infinity. See CL037. In other words, K028 is the locus of the intersection of the line OP and the Steiner line of P when P traverses the circumcircle. See Q011 for an analogous property with the Simson line.

K028 is a stelloid or also a harmonic curve that solves Laplace's equation.

One of the groups of pivots on K028 consists in H (counted twice) and O. In other words, for any point M on K028 and for any arbitrary line L, 2(HM,L)+(OM,L) = constant (mod.π).

It follows that, for any points M and N on K028, we have 2(HM,HN)+(OM,ON) = 0 (mod.π).

Hence, any circle passing through O and H meets the cubic at three other points which are the vertices of an equilateral triangle inscribed in K028.

If M1, M2, M3 are the vertices of an equilateral triangle inscribed in the circumcircle of ABC then the Steiner lines of these points meet the corresponding lines passing through O at P1, P2, P3 and P1P2P3 is an equilateral triangle inscribed in K028 whose circumcircle passes through O and H.

There are infinitely many groups of pivots {P0, P1, P2} on K028 and the circumcircle of P0P1P2 always meets K028 again at the vertices of an equilateral triangle inscribed in K028.

Let P0 be any point on K028 and let Q0 be the homothetic of P0 under h(X381, -1/2). Recall that X381 is the radial center of K028 i.e. the intersection of its asymptotes and then, any line passing through X381 meets K028 at three points whose isobarycenter is X381. This is obviously the case when the line is the Euler line.

The ellipse with foci G, H passing through Q0 is tangent at Q0 to the external bisector at Q0 of triangle GHQ0. The tangents drawn from P0 to this ellipse meet this bisector at two points P1, P2 which are the pivots associated to P0. Note that the ellipse is the Steiner inellipse of triangle P0P1P2.

It follows that, for any two points M, N on K028, we have (P0M,P0N)+(P1M,P1N)+(P2M,P2N) = 0 (mod. π).

A square inscribed in K028

The parallels at O to the nodal tangents (i.e. to the asymptotes of the Jerabek hyperbola) meet the circumcircle at four points M1, M2, M3, M4 and K028 at four points P1, P2, P3, P4 which are the vertices of a square inscribed in K028.

Naturally, each point Pi lies on the Steiner line of Mi.

The cicumcircle of this square is the circle with center O passing through H.

Points on the circumcircle

K028 and K006 meet the circumcircle at the same points namely A, B, C and A', B', C'.

The tangents at A', B', C' to K028 are concurrent at X lying on the lines X(3)-X(1495), X(5)-X(64), X(30)-X(599), X(74)-X(1995), X(110)-X(378), X(125)-X(381), etc. X is X(11472) in ETC.

Recall that the tangents to K006 at these points are concurrent at X(25).

Points on the perpendicular bisectors

K028 meets the perpendicular bisector of BC at O and two other points A1, A2 that lies on the circle HBC. Four other points B1, B2, C1, C2 are defined likewise.

K028 meets the line AO again at A' that lies on the parallel to BC at H. B' and C' are defined likewise.

Fuhrmann triangle and Fuhrmann circle

K028 passes through X(4), X(8), the endpoints of a diameter of the Fuhrmann circle.

Their three remaining common points are the vertices of the Fuhrmann triangle FaFbFc.

These points are the reflections in the sidelines of ABC of the vertices A1, B1, C1 of the circumcevian triangle of the incenter X(1).

Points on the asymptotes

K028 meets its three asymptotes at three collinear points lying on the satellite of the line at infinity. This latter line (S) is parallel to the tangent at O to K028, the line through X(74), X(110), etc.

It meets the Euler line at the homothetic of X(381) under h(H, 1/3), now X(14269) in ETC (2017-09-06).

Hessian, prehessian and flexes of K028

Hessian

The hessian of K028 is a strophoid with node H and singular focus X(381).

Since K028 is a crunodal cubic, it has only three flexes and one only is real, namely F on the figure. Naturally, these flexes also lie on the hessian.

F lies on the perpendicular L at G to the Euler line.

The rectangular hyperbola with center X(110), with asymptotes parallel to those of the Jerabek hyperbola (therefore to the nodal tangents) and passing through X(895) meets the circumcircle at X(74) and three other points, one only being real. The corresponding points on K028 are the flexes.

This strophoid is the inverse in the polar circle of the rectangular hyperbola (H) with center X(468), with asymptotes parallel to those of the Jerabek hyperbola passing through H and X(67). The inverse of L is the circle with diameter HX(468) meeting (H) at the inverse of F.

Prehessian

K028 has only one prehessian. In other words, there is one and only one cubic whose hessian is K028.

This prehessian is also a nodal cubic with node H, passing through X(1656) and obviously the common flexes of K028 and its hessian.